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Question

Find a unit vector perpendicular to each of the vectors a+b and ab where a=3^i+2^j+2^k and b=^i+2^j2^k.

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Solution

Given, a=3^i+2^j+2^k

and b=^i+2^j2^k.

Now, a+b=(3^i+2^j+2^k)+(^i+2^j2^k)=4^i+4^j+0^k

ab=(3^i+2^j+2^k)(^i+2^j2^k)=2^i+0^j+4^k

We know that the unit vector perpendicular to both a+b ane ab is given by

^n=(a+b)×(ab)(a+b)×(ab)

Here, (a+b)×(ab)=∣ ∣ ∣^i^j^k440204∣ ∣ ∣

=^i(160)^j(160)+^k(08)

=16^i16^j8^k

Now, (a+b)×(ab)=(16)2+(16)2+(8)2

=256+256+64=576=24

Hence, the required unit vector ^n=(a+b)×(ab)(a+b)×(ab)

=16^i16^j8^k24=23^i23^j13^k

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